How many different 6-letter sequences are there that consist

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How many different 6-letter sequences are there that consist [#permalink]

28 Jun 2010, 10:48

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Question Stats:

37% (01:26) correct

62% (00:56) wrong

based on 3 sessions

How many different 6-letter sequences are there that consist of 1A, 2B's and 3C's ?

A. 6
B. 60
C. 120
D. 360
E. 720

I thought the answer is 6!/2!3! = 60

But the answer is not 60.

[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Mar 2013, 09:21, edited 1 time in total.

Renamed the topic and edited the OA.

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Re: Counting -- [#permalink]

28 Jun 2010, 11:02

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testprep2010 wrote:

Please Solve:

How many different 6-letter sequences are there that consist of 1A, 2B's and 3C's ?

A 6
B 60
C 120
D 360
E 720

I thought the answer is 6!/2!3! = 60

But the answer is not 60.

# of different permutations of 6 letters ABBCCC is indeed \frac{6!}{2!3!}=60, so you are right answer must be B (60).

THEORY:
Permutations of n things of which P_1 are alike of one kind, P_2 are alike of second kind, P_3 are alike of third kind ... P_r are alike of r_{th} kind such that: P_1+P_2+P_3+..+P_r=n is:

\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \frac{6!}{2!2!}, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \frac{9!}{4!3!2!}.
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Re: Counting -- [#permalink]

28 Jun 2010, 11:05

The answer should be 6!/2!3! = 60
What's the source?

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Re: Counting -- [#permalink]

28 Jun 2010, 11:15

Bunuel , prabir2001 : Thanks for the solution.

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Re: Counting -- [#permalink]

28 Jun 2010, 15:51

great post, bunuel!

Here's an OG12 question where one can apply this concept: page 179, question 191.
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6-letter sequences [#permalink]

15 Nov 2010, 07:54

How many different 6-letter sequences are there that consist of 1 A, 2B's, and 3 C's?
(A) 6
(B) 60
(C) 120
(D) 360
(E) 720
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Re: Counting -- [#permalink]

19 Mar 2013, 09:12

There is another approach as we can arrange from either A to C or C to A. So...

6C1 X 5C2 (A to C) = 60; It is equal to 6C3 X 3C2 (C to A) which has the same result.

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Re: Counting -- [#permalink]

19 Mar 2013, 09:20

curtis0063 wrote:

There is another approach as we can arrange from either A to C or C to A. So...

6C1 X 5C2 (A to C) = 60; It is equal to 6C3 X 3C2 (C to A) which has the same result.

Yup this question can be solved a couple of different ways. The easiest is obviously the permutation formula for repeating elements quoted above by Bunuel. The answer is 60, regardless of what the OA indicates.

This highlights a recurring theme on GMAT problems. You should always double check the math for yourself and not trust the material blindly. Typos happen, mistakes occur. If you understand how to solve the questions, you'll get them right on the GMAT as that material is checked, rechecked and checked again for mistakes.

Thanks!
-Ron
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Re: Counting -- [#permalink] 19 Mar 2013, 09:20

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